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It is common for test-takers to think in the right direction, understand what a question gives and what it is asking to be found out, but still get the wrong answer. Mistakes made during the execution of a problem are common on exams, but what is rather rare is going with incorrect logic and still getting the correct answer! If only life was this rosy so often!

Today, we will look at a question in which exactly this phenomenon occurs – we will find the flaw in the logic that test-takers often come up with and then learn how to correct that flaw:

If a motorist had driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster, he would have covered 70 more miles than he actually did. How many more miles would he have covered than he actually did had he driven 2 hours longer and at an average rate of 10 miles per hour faster on that day?

(A) 100

(B) 120

(C) 140

(D) 150

(E) 160

**Method 1:** Algebra

Let’s start with the basic “Distance = Rate × Time” formula:

……….(I)

From here, the first theoretical trip can be represented as , (the motorist travels for 1 extra hour at a rate of 5 mph faster), which can be expanded to .

We can then eliminate “” by plugging in the value of “” from our equation (I):

, which simplifies to and then to ……….. (II)

The second theoretical trip can be represented as , which expands to (not that we only have an expression since we don’t know what the distance is).

The two middle terms can be factored to , which allows us to use equation (II) here:

.

Since the original distance was , the additional distance is 150 more miles, or answer choice (D).

We totally understand that this solution is a bit convoluted – algebra often is. So, understandably, students often look for a more direct logical solution.

Here is one they sometimes employ:

**Method 2:** Logic (Incorrect)

If the motorist had driven 1 hour longer at a rate 5 mph faster, then his original speed would be 70 miles subtracted by the extra 5 miles he drove in that hour to get 70 – 5 = 65 mph. If he drives at a rate 10 mph faster (i.e. at 65 + 10 = 75) × 2 for the extra hours, he/she would have driven 150 miles extra.

But here is the catch in this logic:

The motorist drove for an average rate of 5 mph extra. So the 70 includes not only the extra distance covered in the last hour, but also the extra 5 miles covered every hour for which he drove. Hence, his original speed is not 65. Now, let’s see the correct logical method of solving this:

**Method 3:** Logic (Correct)

Let’s review the original problem first. Say, speed is “” mph – we don’t know the number of hours for which this speed was maintained.

STEP 1:

= TOTAL DISTANCE COVERED

In the first hypothetical case, the motorist drove for an extra hour at a speed of 5 mph faster. This means he covered 5 extra miles every hour and then covered another + 5 miles in the last hour. The underlined distances are the extra ones which all add up to 70.

STEP 2:

= TOTAL DISTANCE COVERED

+5 +5 +5 + … + 5 + 5 = +70

In the second hypothetical case, in which the motorist drove for two hours longer at a speed of 10 mph faster, he adds another 5 mph to his hourly speed and covers yet another distance of “” in the second extra hour. In addition to S, he also covers another 10 miles in the second extra hour. The additional distances are shown in red in the third case – every hour, the speed is 10 mph faster and he drove for two extra hours in this case (compared with Step 1).

STEP 3:

= TOTAL DISTANCE COVERED

+5 +5 +5 + … +5 +5 +5 = +70

+5 +5 +5 + … +5 +5 +5 + 10 = +70 + 10

Note that the +5s and the all add up to 70 (as seen in Step 2). We also separately add the extra 10 from the last hour. This is the logic of getting the additional distance of 70 + 70 + 10 = 150. It involves no calculations, but does require you to understand the logic.

Therefore, our answer is still **(D)**.